Evaluate the limit: $\lim_{x\to 0} x.\sin \left(\dfrac {1}{x} \right)$
My Attempt:
We know, $$-1\leq \sin \left(\dfrac {1}{x}\right) \leq 1$$ Multiplying each term by $x$ $$-x\leq x \sin \left(\dfrac {1}{x}\right) \leq x$$ for $x>0$
$$-x\geq x \sin \left(\dfrac {1}{x}\right) \geq x$$ for $x<0$
You have the right approach, this is the canonical example of the squeeze theorem.
Take care to approach zero from both negative and positive sides and conclude that their limits are the same.